Solving nonhomogeneous ordinary differential equation by use of Laplace transform

Question

Given the differential equation $$ ′′ − 5′ + 6 = e^{−}, \quad (0) = _1, \, ’(0) = _2 $$ how can I use the Laplace transform to solve this nonhomogeneous differential equation ?

Answer 1

The essential method is to take the Laplace transform of both sides of the differential equation as seen by the following. \begin{align} \mathcal{L}\{ y'' - 5 \, y' + 6 \, y \} &= \mathcal{L}\{ e^{-a t} \} \\ s^2 \, f(s) - s \, y(0) - y'(0) - 5 \, s \, f(s) + 5 \, y(0) + 6 \, f(s) &= \frac{1}{s+a} \\ (s^2 - 5 \, s + 6) \, f &= \frac{1}{s+a} + (s-5) \, y(0) + y'(0) \end{align} This leads to $$f = \frac{1}{(s+a)(s-2)(s-3)} + \frac{(s-5)}{(s-2)(s-3)} \, y(0) + \frac{y'(0)}{(s-2)(s-3)}. $$ Inversion gives \begin{align} y(t) &= \frac{e^{-a t}}{(a+2)(a+3)} + \left(\frac{1}{a+3} - 2 \, y(0) + y'(0) \right) \, e^{3 t} - \left(\frac{1}{a+2} - 3 \, y(0) + y'(0) \right) \, e^{2 t}. \end{align}

Using the initial condition and the chosen value for $a$ gives the desired result. Also note that the Laplace transform used is defined by $$ f(s) = \int_{0}^{\infty} e^{- s t} \, y(t) \, dt. $$